Solve 36+sqrt{r^2-36}=r^2

$$36 + \sqrt { r ^ { 2 } - 36 } = r ^ { 2 }$$

Solve for r

$$r=\sqrt{37}\approx 6.08276253$$

$$r=-\sqrt{37}\approx -6.08276253$$

$$r=-6$$

$$r=6$$

Solution Steps

Subtract 36 from both sides of the equation.

$$\sqrt{r^{2}-36}=r^{2}-36$$

Square both sides of the equation.

$$\left(\sqrt{r^{2}-36}\right)^{2}=\left(r^{2}-36\right)^{2}$$

Calculate \sqrt{r^{2}-36} to the power of 2 and get r^{2}-36.

$$r^{2}-36=\left(r^{2}-36\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(r^{2}-36\right)^{2}.

$$r^{2}-36=\left(r^{2}\right)^{2}-72r^{2}+1296$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$r^{2}-36=r^{4}-72r^{2}+1296$$

Subtract r^{4} from both sides.

$$r^{2}-36-r^{4}=-72r^{2}+1296$$

Add 72r^{2} to both sides.

$$r^{2}-36-r^{4}+72r^{2}=1296$$

Combine r^{2} and 72r^{2} to get 73r^{2}.

$$73r^{2}-36-r^{4}=1296$$

Subtract 1296 from both sides.

$$73r^{2}-36-r^{4}-1296=0$$

Subtract 1296 from -36 to get -1332.

$$73r^{2}-1332-r^{4}=0$$

Substitute t for r^{2}.

$$-t^{2}+73t-1332=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute -1 for a, 73 for b, and -1332 for c in the quadratic formula.

$$t=\frac{-73±\sqrt{73^{2}-4\left(-1\right)\left(-1332\right)}}{-2}$$

Do the calculations.

$$t=\frac{-73±1}{-2}$$

Solve the equation t=\frac{-73±1}{-2} when ± is plus and when ± is minus.

$$t=36 t=37$$

Since r=t^{2}, the solutions are obtained by evaluating r=±\sqrt{t} for each t.

$$r=6 r=-6 r=\sqrt{37} r=-\sqrt{37}$$

Substitute 6 for r in the equation 36+\sqrt{r^{2}-36}=r^{2}.

$$36+\sqrt{6^{2}-36}=6^{2}$$

Simplify. The value r=6 satisfies the equation.

$$36=36$$

Substitute -6 for r in the equation 36+\sqrt{r^{2}-36}=r^{2}.

$$36+\sqrt{\left(-6\right)^{2}-36}=\left(-6\right)^{2}$$

Simplify. The value r=-6 satisfies the equation.

$$36=36$$

Substitute \sqrt{37} for r in the equation 36+\sqrt{r^{2}-36}=r^{2}.

$$36+\sqrt{\left(\sqrt{37}\right)^{2}-36}=\left(\sqrt{37}\right)^{2}$$

Simplify. The value r=\sqrt{37} satisfies the equation.

$$37=37$$

Substitute -\sqrt{37} for r in the equation 36+\sqrt{r^{2}-36}=r^{2}.

$$36+\sqrt{\left(-\sqrt{37}\right)^{2}-36}=\left(-\sqrt{37}\right)^{2}$$

Simplify. The value r=-\sqrt{37} satisfies the equation.

$$37=37$$

List all solutions of \sqrt{r^{2}-36}=r^{2}-36.

$$r=6 r=-6 r=\sqrt{37} r=-\sqrt{37}$$